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(15+2x)(10+2x)=360
We move all terms to the left:
(15+2x)(10+2x)-(360)=0
We add all the numbers together, and all the variables
(2x+15)(2x+10)-360=0
We multiply parentheses ..
(+4x^2+20x+30x+150)-360=0
We get rid of parentheses
4x^2+20x+30x+150-360=0
We add all the numbers together, and all the variables
4x^2+50x-210=0
a = 4; b = 50; c = -210;
Δ = b2-4ac
Δ = 502-4·4·(-210)
Δ = 5860
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5860}=\sqrt{4*1465}=\sqrt{4}*\sqrt{1465}=2\sqrt{1465}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-2\sqrt{1465}}{2*4}=\frac{-50-2\sqrt{1465}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+2\sqrt{1465}}{2*4}=\frac{-50+2\sqrt{1465}}{8} $
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